To determine the number of metals that will be oxidized by
Cr2O72− in aqueous solution, we need to compare their standard reduction potentials with that of the given reduction potential of
Cr2O72− .
The reduction reaction for
Cr2O72− is:
Cr2O72−+14H++6e−→2Cr3++7H2O,E∘=1.33 V
This means that
Cr2O72− has a strong tendency to get reduced (due to its high reduction potential of 1.33 V). Thus, any metal with a lower standard reduction potential than 1.33 V has the potential to be oxidized by
Cr2O72− .
Given the standard reduction potentials:
Fe3+(aq)+3e−→FeNi2+(aq)+2e−→NiAg+(aq)+e−→AgAu3+(aq)+3e−→AuE∘=−0.04 VE∘=−0.25 VE∘=0.80 VE∘=1.40 V
Now we compare these reduction potentials with that of
Cr2O72− :
- For
Fe3+/Fe , E∘=−0.04 V (lower than 1.33 V)
- For
Ni2+/Ni , E∘=−0.25 V (lower than 1.33 V)
- For
Ag+/Ag , E∘=0.80 V (lower than 1.33 V)
- For
Au3+/Au , E∘=1.40 V (higher than 1.33 V)
From the comparison, we see that
,
, and
have standard reduction potentials lower than 1.33 V, meaning these metals can be oxidized by
Cr2O72− , but
cannot.
Therefore, the number of metals that will be oxidized by
Cr2O72− in aqueous solution is:
3.
